# Do I have a problem? Basic proportion problem.

1. Aug 11, 2012

### Life-Like

Hello, this is an odd question but when I compared how I solved it compared to how the book solved it I kinda got worried. It was a very basic problem I been going back and touching up on my basics before I start pre-cal and trig in college.

The problem was: After completing 7/10 of his math homework assignment, Josh has 15 more questions complete. What is the total number of questions on his assignment? The answer was 50.

I got that by

$\frac{7}{10}$=$\frac{(x-15)}{x}$ Followed by cross multiplying and dividing.

However the book does it simply by doing the subtraction 10-7 = 3 then setting it up as

$\frac{3}{10}$=$\frac{15}{x}$ Then cross multiply and divide.

I know someone will say "It works so you can do what you are doing". But is there a possibility this kind of over complication( haha not hardly complicated!) Can hurt me later on? The way I performed it seemed much more intuitive... I have a highschool record of up to ap calc and ap chem and I'm worried if I pursue engineering in college I'll have a hard time solving simple things because of my use to complexity?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 11, 2012

### azizlwl

(3/10)x=15
x=50

3. Aug 11, 2012

### HallsofIvy

Staff Emeritus
Do you meant "15 more questions to complete"? What you wrote could be interpreted as meaning has completed 15 questions. If he has already completed 7/10 of his assignment, he still has 3/10 left. If he has 15 questions more to complete, letting "x" be the total number of problems, (3/10)x= 15 so x= (10/3)(15)= 150/3= 50.

Okay, that is, they are calculating 1- 7/10= 10/10- 7/10= (10- 7)/10= 3/10 as I did.

As long as you have linear problems, you can set them up as "proportions" and get the correct answer.

4. Aug 11, 2012

### Life-Like

Thanks guys

Last edited: Aug 11, 2012